Next question: A garden hose with internal diameter of 13.5 mm lies flat on a sidewalk while water is flowing in it at a speed of 6 m/s. A person happens to step on it at the very edge of the opening of the hose and decreases its internal diameter by a factor of 9 So D (1) = 0.0135m r (1) = 0.00675m D (2) = 0.0135/9 = 0.0015m r (2) = 0.00075m A (1) = pi*r^2 = (3.14...)(0.00675^2) = 1.4134...e-4 A (2) = pi*r^2 = (3.14...)(0.00075^2) = 1.767...e-6 1. What was the water flow rate in the hose prior to the person stepping on it? - I got this part: J(1) = A(1)V(1) = 8.59 m3/s 2. What is the flow rate of water after the person steps on it? 3. What is the speed of the water just as it exits the hose after the person steps on it?